= Solution
Use $\mu=\gamma m$ for the gravitational parameter, to distinguish it from the Delaunay action called $G$. The displayed actions are per unit reduced mass. Accordingly use specific canonical momenta; with physical relative momenta, all three actions acquire a common reduced-mass factor, which leaves the eccentricity fraction below unchanged.
\b[The canonical transformation.] In spherical coordinates, separation of the Kepler <Hamilton-Jacobi equation> gives
$$
p_\phi=H,\qquad p_\theta^2=G^2-\frac{H^2}{\sin^2\theta},\qquad p_r^2=2E+\frac{2\mu}{r}-\frac{G^2}{r^2},\qquad E=-\frac{\mu^2}{2L^2}.
$$
A local type-two generating function, continued across turning points with the appropriate momentum branches, is
$$
W=H\phi+\int^\theta\sqrt{G^2-H^2/\sin^2\theta'}\,d\theta'+\int^r\sqrt{-\mu^2/L^2+2\mu/r'-G^2/r'^2}\,dr'.
$$
The old momenta are $\partial W/\partial(r,\theta,\phi)$ and the new angles are $l=\partial_LW$, $g=\partial_GW$, $h=\partial_HW$. This defines a <canonical transformation>, since its generating differential is $dW=p_rdr+p_\theta d\theta+p_\phi d\phi+l\,dL+g\,dG+h\,dH$. With standard choices of angle origins, the resulting <Delaunay variables> are
$$
L=\sqrt{\mu a},\qquad G=L\sqrt{1-e^2},\qquad H=G\cos i,
$$
with $l$ the <mean anomaly>, $g$ the <argument of periapsis>, and $h$ the <longitude of ascending node>. In particular, the signed $H$ includes retrograde orbits; the squared equation in the PDF must not be read as $H\geq0$.
For bound noncollision <Kepler orbits>, the independent ranges are
$$
\boxed{0<L<\infty,\quad0<G<L,\quad-G<H<G,\quad0\leq l,g,h<2\pi.}
$$
The limiting circular, radial and coplanar cases sit on boundaries where some angle charts degenerate. They have zero measure in the generic phase-volume calculation. The ranges are constrained as above, rather than a Cartesian box with three unconstrained actions.
\b[The eccentricity count.] A <canonical transformation> preserves <phase space> volume, so
$$
dN=F(E(L))\,dL\,dG\,dH\,dl\,dg\,dh.
$$
Integrating the three angles contributes $(2\pi)^3$, and integrating $H$ gives $2G$. At fixed $L$, the condition that the <orbital eccentricity> be less than $e_0$ is $G>L\sqrt{1-e_0^2}$. Thus
$$
\int_{L\sqrt{1-e_0^2}}^L2G\,dG=L^2e_0^2,\qquad \int_0^L2G\,dG=L^2.
$$
Any energy weight cancels between these integrals. If the total population is finite, or an explicit finite action band is used,
$$
\boxed{N(e<e_0)=e_0^2N_{\rm tot},\qquad P(e<e_0)=e_0^2,\qquad f_e(e)=2e.}
$$
This is the <Delaunay phase-volume proof of the thermal eccentricity distribution>. The source's $e^2$ law is a cumulative count, not a differential eccentricity density. Also, pointwise finite $F(E)$ does not alone imply finite $N_{\rm tot}$: $F=1$ gives $\int_0^\infty L^2dL=\infty$. The <finite normalization of an energy-only Kepler distribution> is the additional condition needed for a global probability statement; the fraction at every fixed energy is unaffected.
\b[Why the observation does not imply relaxation or equilibrium.] No <Boltzmann distribution> was used: every normalizable energy-only <phase-space distribution function> has the same eccentricity marginal, irrespective of collisional thermalization. An even stronger counterexample is a genuinely time-dependent collisionless distribution
$$
f(L,G,H,l,g,h,t)=F(E(L))[1+\eta\cos(l-\nu(L)t)],\qquad0<|\eta|<1,\qquad\nu(L)=\frac{\mu^2}{L^3}.
$$
Kepler motion has $\dot l=\nu(L)$ and constant actions and other angles, so $\partial_tf+\nu\partial_lf=0$: this is an exact nonstationary solution of the <Collisionless Boltzmann equation>. Integrating $l$ removes the cosine and leaves precisely the same $e^2$ cumulative count at every time. \b[An eccentricity marginal alone cannot establish thermal relaxation, a relaxed age, or even a stationary distribution.]
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